Christopher Barker wrote:
> It does, but we don't need a binary installer for a python that doesn't
> have a binary installer.
>
Yes, not now - but I would prefer avoiding to have to change the process
again when time comes. It may not look like it, but enabling a working
process which works w
Robert Kern gmail.com> writes:
> On Sat, Mar 28, 2009 at 23:15, Anne Archibald gmail.com>
wrote:
> > 2009/3/28 Geoffrey Irving naml.us>:
> >> On Sat, Mar 28, 2009 at 12:47 AM, Robert Kern gmail.com>
wrote:
> >>> 2009/3/27 Charles R Harris gmail.com>:
>
> On Fri, Mar 27, 2009 at 4:43
Same. Thanks, too.
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> The array follows a pattern: each array of length 2 represents the x,y
index of that array within the larger array.
Is this what you are after?
>>> numpy.array(list(numpy.ndindex(n,n))).reshape(n,n,2)
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Thanks!
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2009/3/31 Ian Mallett :
> The array follows a pattern: each array of length 2 represents the x,y index
> of that array within the larger array.
Ah, right. Use dstack(mgrid[0:n,0:n]).
--
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terribl
The array follows a pattern: each array of length 2 represents the x,y index
of that array within the larger array.
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2009/3/31 Ian Mallett :
> On Tue, Mar 31, 2009 at 5:32 PM, Robert Kern wrote:
>>
>> How do you want to fill in the array? If you are typing it in
>> literally into your code, you would do basically the above, without
>> the ...'s, and wrap it in numpy.array(...).
>
> I know that, but in some cases
On Tue, Mar 31, 2009 at 5:32 PM, Robert Kern wrote:
> How do you want to fill in the array? If you are typing it in
> literally into your code, you would do basically the above, without
> the ...'s, and wrap it in numpy.array(...).
I know that, but in some cases, n will be quite large, perhaps 1
2009/3/31 Ian Mallett :
> Hello,
> I'm trying to make an array of size n*n*2. It should be of the form:
> [[[0,0],[1,0],[2,0],[3,0],[4,0], ... ,[n,0]],
> [[0,1],[1,1],[2,1],[3,1],[4,1], ... ,[n,1]],
> [[0,2],[1,2],[2,2],[3,2],[4,2], ... ,[n,2]],
> [[0,3],[1,3],[2,3],[3,3],[4,3], ... ,[n,3]],
>
Hello,
I'm trying to make an array of size n*n*2. It should be of the form:
[[[0,0],[1,0],[2,0],[3,0],[4,0], ... ,[n,0]],
[[0,1],[1,1],[2,1],[3,1],[4,1], ... ,[n,1]],
[[0,2],[1,2],[2,2],[3,2],[4,2], ... ,[n,2]],
[[0,3],[1,3],[2,3],[3,3],[4,3], ... ,[n,3]],
[[0,4],[1,4],[2,4],[3,4],[4,4], ... ,
David Cournapeau wrote:
> Chris Barker wrote:
>> Well, neither Apple nor python.org's builds are 64 bit anyway at this
>> point. There is talk of quad (i386,and ppc_64 i86_64) builds the the
>> future, though.
>>
> Yes, but that's something that has to should be supported sooner rather
> than
>
> I think dot will work, though you'll need to work a little bit to get the
> answer:
>
> >>> import numpy as np
> >>> a = np.array([[1,2], [3,4]], np.float)
> >>> aa = np.array([a,a+1,a+2])
> >>> bb = np.array((a*5, a*6, a*7, a*8))
> >>> np.dot(aa, bb).shape
> (3, 2, 4, 2)
> >>> for i, a_ in
Hi David, Stefan,
The windows buildbot is back online but seems to have a configuration
problem. It would be nice to see that build working before the release, so
it would be nice if you two could take a look at the error messages/contact
Heller.
Chuck
Pauli Virtanen wrote:
>
> Probably they are both related to unspecified sort order for
> the duplicates. There were some sort-order ignoring missing in the test.
>
> I think the test is now fixed in trunk:
>
> http://projects.scipy.org/numpy/changeset/6827
>
The test passes in 1.4.0.dev68
Le Friday 27 March 2009 23:38:25 Bryan Cole, vous avez écrit :
> I have a number of arrays of shape (N,4,4). I need to perform a
> vectorised matrix-multiplication between pairs of them I.e.
> matrix-multiplication rules for the last two dimensions, usual
> element-wise rule for the 1st dimension (
Charles R Harris wrote:
> That should work. The first two integrals are actually the same, but
> need to be E(t)*E(t).conj(). The second integral needs twice the real
> part of E(t)*E(t-tau).conj(). Numpy correlate should really have the
> conjugate built in, but it doesn't.
>
> Chuck
>
It wo
On Tue, Mar 31, 2009 at 8:54 PM, Jochen S wrote:
> On Tue, Mar 31, 2009 at 7:13 AM, João Luís Silva wrote:
>
>> Hi,
>>
>
>
>> I wrote a script to calculate the *optical* autocorrelation of an
>> electric field. It's like the autocorrelation, but sums the fields
>> instead of multiplying them. I'
On Tue, Mar 31, 2009 at 7:13 AM, João Luís Silva wrote:
> Hi,
>
> I wrote a script to calculate the *optical* autocorrelation of an
> electric field. It's like the autocorrelation, but sums the fields
> instead of multiplying them. I'm calculating
>
> I(tau) = integral( abs(E(t)+E(t-tau))**2,t=
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